Question

Two forces P and Q have a resultant R.This resultant is doubled,when Q is doubled,or when Q is reversed. Show that, P:Q:R=
$\sqrt{2 \: } \binom{.}{.} \sqrt{3} \binom{.}{.} \sqrt{2}$


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Answer 1

Answer:

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Answer 2

R² = P² + Q² + 2PQcosθ …. (1)

where θ is the angle between P and Q

Now, if Q is Doubled the new resultant S is perpendicular to P

=> cos 90 = ( P + 2Qcosθ ) / S

=> 0 = P + 2Qcosθ

=> cosθ =- P/2Q …… (2)

Putting Value of (2) in (1), we get

R² = P² + Q² + 2PQ(-P/2Q)

=> R² = P² + Q² - P²

=> R² = Q² Hence R = Q